Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $x = \dfrac{t - 2}{t + 5} \div \dfrac{t^2 + 2t - 8}{4t + 20} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{t - 2}{t + 5} \times \dfrac{4t + 20}{t^2 + 2t - 8} $ First factor the quadratic. $x = \dfrac{t - 2}{t + 5} \times \dfrac{4t + 20}{(t - 2)(t + 4)} $ Then factor out any other terms. $x = \dfrac{t - 2}{t + 5} \times \dfrac{4(t + 5)}{(t - 2)(t + 4)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (t - 2) \times 4(t + 5) } { (t + 5) \times (t - 2)(t + 4) } $ $x = \dfrac{ 4(t - 2)(t + 5)}{ (t + 5)(t - 2)(t + 4)} $ Notice that $(t + 5)$ and $(t - 2)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 4\cancel{(t - 2)}(t + 5)}{ (t + 5)\cancel{(t - 2)}(t + 4)} $ We are dividing by $t - 2$ , so $t - 2 \neq 0$ Therefore, $t \neq 2$ $x = \dfrac{ 4\cancel{(t - 2)}\cancel{(t + 5)}}{ \cancel{(t + 5)}\cancel{(t - 2)}(t + 4)} $ We are dividing by $t + 5$ , so $t + 5 \neq 0$ Therefore, $t \neq -5$ $x = \dfrac{4}{t + 4} ; \space t \neq 2 ; \space t \neq -5 $